Find the equation of the tangent line

Question

Find the equation of the tangent line to the graph of the equation at the given point:

ln(10 - y²) + 16 = -2x³ at (-2, 3)

Answer 1

Hi Stormee,

We can write the equation of the tangent line in point-slope form, which is y - y₀ = m(x - x₀), where (x₀, y₀) is the point given to us and m is the slope (or derivative dy/dx) at this point. We are given the point (-2, 3), so now we have y - 3 = m(x + 2).

To find the derivative at this point, we first have to find dy/dx of this function using implicit differentiation with respect to x. The work is shown below.

d/dx [ln(10 - y²) + 16] = d/dx [-2x³]

⇒ d/dx[10 - y²] * 1 / (10 - y²) = -6x² (the left-hand side is the chain rule used on ln. Remember that d/dx [ln(x)] = 1/x.

⇒ -2y * dy/dx * 1 / (10 - y²) = -6x² (The resulting dy/dx is also a result of the chain rule).

Solving the above equation for the slope dy/dx, we get

dy/dx = 3x²(10 - y²) / y.

To find the derivative at the given point (-2, 3), we plug in x = -2 and y = 3 into dy/dx, getting

dy/dx at (-2, 3) = m = 3(-2)²(10 - 3²) / 3 = 4

Substituting this into the equation for the line, we have our answer of

y - 3 = 4(x + 2)

or in slope-intercept form,

y = 4x + 11

Answer 2

implicit differentiation:

(-2y)y'/(10-y^2) = -6x^2

(-2y y') = (-6x^2)(10 - y^2)

y' = (-6x^2)(10 - y^2) / (-2y)

= (3x^2)(10 - y^2)/y

For the given x=-2 and y=3

the slope is M = (3*(-2)*(-2)) ( 10 - (3)^2) / (3)

= (12 * (10 - 9))/3

= 12/3

= 4

So the tangent line has slope 4 and eats(-2,3)

B = y - mx = 3 - (4)(-2)

= 3 - -8

= 11

y = 4x+11 is the tangent line